Define-syntax scheme usage

Question

since yesterday I've been trying to program a special case statement for scheme that would do the following:

(define (sort x)
  (cond ((and (list? x) x) => (lambda (l)
                                (sort-list l)))
        ((and (pair? x) x) => (lambda (p)
                        (if (> (car p) (cdr p))
                            (cons (cdr p) (car p))
                            p)))
        (else "here")))

instead of using all the and's and cond's statement, I would have:

(define (sort x)
  (scase ((list? x) => (lambda (l)
                                (sort-list l)))
         ((pair? x) => (lambda (p)
                        (if (> (car p) (cdr p))
                            (cons (cdr p) (car p))
                            p)))
        (else "here")))

What I could do so far, was this:

(define (sort x)
  (scase (list? x) (lambda (l)
                      (sort-list l)))
  (scase (pair? x) (lambda (p)
                        (if (> (car p) (cdr p))
                            (cons (cdr p) (car p))
                            p))))

with this code:

(define-syntax scase
  (syntax-rules ()
    ((if condition body ...)
     (if condition
         (begin
           body ...)))))

What I wanted to do now, is just allow the scase statement to have multiple arguments like this:

(scase ((list? (cons 2 1)) 'here)
       ((list? '(2 1)) 'working))

but I can't seem to figure out how I can do that. Maybe you guys could give me a little help?

Thanks in advance ;)

Theo Belaire · Accepted Answer · 2012-07-03 17:57:12Z

If this is an exercise in learning how to use syntax-rules, then disregard this answer.

I see a way to simplify your code that you are starting with.

(define (sort x)
  (cond ((list? x)
            (sort-list x))
        ((pair? x)
            (if (> (car x) (cdr x))
                (cons (cdr x) (car x))
                x)))
        (else "here")))

Since all the (and (list? x) x) => (lambda l ... does is see if x is a list, and then bind l to x, (since #f is not a list, and '() is not false, at least in Racket), you can just skip all that and just use x. You do not need to use => in case, and in this case it doesn't help. => is useful if you want to do an test that returns something useful if successful, or #f otherwise.

Now, if you want to use a macro, then you're going to need to clarify what you want it to do a bit better. I think that case already does what you want. Your existing macro is just if, so I'm not sure how to extend it.

João Amaral · Accepted Answer · 2012-07-04 00:30:02Z

2

I found the solution for my question, here it goes:

(define-syntax cases
  (syntax-rules ()
    ((_ (e0 e1 e2 ...)) (if e0 (begin e1 e2 ...)))
    ((_  (e0 e1 e2 ...) c1 c2 ...)
     (if e0 (begin e1 e2 ...) (cases c1 c2 ...)))))

Thank you all anyway :)

answered Jul 4, 2012 at 0:30

João Amaral

895 bronze badges

Comments

Rajesh Bhat · Accepted Answer · 2012-07-04 02:24:08Z

0

Here's a solution :

#lang racket

(require mzlib/defmacro)

(define-syntax scase
  (syntax-rules (else)
    ((_ (else body1)) body1)
    ((_ (condition1 body1) (condition2 body2) ...)
     (if condition1
         body1
         (scase (condition2 body2) ...)))))

(define (sort1 x)
  ((scase ((list? x) (lambda (l)
                      (sort l <)))
         ((pair? x) (lambda (p)
                      (if (> (car p) (cdr p))
                          (cons (cdr p) (car p))
                          p)))
         (else (lambda (e) "here")))
   x))

It works in DrRacket. I made three changes to your solution. First, i renamed your sort procedure to sort1 since sort is inbuilt in scheme ( I have used it inside sort1). Second, I have changed the sort1 itself so that the input given will be passed to the procedure returned by scase and you will directly get the sorted result. Third, I have modified the scase syntax extension, so that it will accept the else condition.

>(sort1 (list 3 1 2))
'(1 2 3)

> (sort1 (cons 2 1))
'(1 . 2)

> (sort1 'here)
"here"

I suggest you read "The Scheme Programming Language" by Kent Dybvig. There is an entire chapter on syntactic extensions.

answered Jul 4, 2012 at 2:24

Rajesh Bhat

1,0006 silver badges8 bronze badges

1 Comment

Theo Belaire Over a year ago

Err, all you've done is re-write cond (without the => "feature"). The code works if you replace scase with cond in your final answer. If that's what you are doing, just say so.

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